Volumes of revolution about the y-axis Problem set B Worked solutions

The part of the curve \(y=3x^{2}+1\) for \(0\le x\le 2\) is rotated 360° about the y-axis. Find the volume of the solid enclosed by the curve, the y-axis and the horizontal lines through the two ends of the curve.

• \(y=3x^{2}+1\Rightarrow x^{2}=\frac{y-1}{3}\)
• Limits: \(x=0\) gives \(y=1\), and \(x=2\) gives \(y=13\)
• \(V=\pi\int_{1}^{13}\frac{y-1}{3}\,dy\)
• From the calculator: \(\pi\int_{1}^{13}\frac{y-1}{3}\,dy = 75.4\)

The part of the curve \(y=\ln x-1\) for \(2\le x\le 5\) is rotated 360° about the y-axis. Find the volume of the solid enclosed by the curve, the y-axis and the horizontal lines through the two ends of the curve.

• \(y=\ln x-1\Rightarrow \ln x=y+1\Rightarrow x=e^{y+1}\), so \(x^{2}=e^{2y+2}\)
• Limits: \(x=2\) gives \(y=\ln 2-1\), and \(x=5\) gives \(y=\ln 5-1\)
• \(V=\pi\int_{\ln 2-1}^{\ln 5-1}e^{2y+2}\,dy\)
• From the calculator: \(\pi\int_{\ln 2-1}^{\ln 5-1}e^{2y+2}\,dy = 33.0\)

The part of the curve \(y=\cos x\) for \(0\le x\le 1.2\) is rotated 360° about the y-axis. Find the volume of the solid enclosed by the curve, the y-axis and the horizontal lines through the two ends of the curve. Use radians.

• \(y=\cos x\Rightarrow x=\cos^{-1}y\), so \(x^{2}=\left(\cos^{-1}y\right)^{2}\)
• Limits: \(x=0\) gives \(y=1\), and \(x=1.2\) gives \(y=\cos 1.2=0.362\)
• Radian mode. The lower y-value goes at the bottom of the integral.
• From the calculator: \(\pi\int_{\cos 1.2}^{1}\left(\cos^{-1}y\right)^{2}dy = 1.38\)

The part of the curve \(y=\frac{3}{x-1}\) for \(2\le x\le 7\) is rotated 360° about the y-axis. Find the volume of the solid enclosed by the curve, the y-axis and the horizontal lines through the two ends of the curve.

• \(y=\frac{3}{x-1}\Rightarrow x-1=\frac{3}{y}\Rightarrow x=\frac{3}{y}+1\)
• \(x^{2}=\left(\frac{3}{y}+1\right)^{2}=\frac{9}{y^{2}}+\frac{6}{y}+1\)
• Limits: \(x=2\) gives \(y=3\), and \(x=7\) gives \(y=\frac{1}{2}\)
• From the calculator: \(\pi\int_{0.5}^{3}\left(\frac{3}{y}+1\right)^{2}dy = 88.8\)

The diagram shows the region bounded by the curve \(y=\frac{3}{x}\), the y-axis and the lines \(y=2\) and \(y=5\).

1. (a) Find the area of the region.

   • \(x=\frac{3}{y}\)
   • Area = \(\int_{2}^{5}\frac{3}{y}\,dy\)
   • From the calculator: \(\int_{2}^{5}\frac{3}{y}\,dy = 2.75\)

2. (b) The region is rotated 360° about the y-axis. Find the exact volume of the solid formed.

   • \(x^{2}=\frac{9}{y^{2}}\)
   • \(V=\pi\int_{2}^{5}\frac{9}{y^{2}}\,dy=\pi\left[-\frac{9}{y}\right]_{2}^{5}\)
   • \(V=\frac{27\pi}{10}\)

The region between the curve \(y=\sqrt{x^{2}+9}\), the y-axis and the line \(y=5\) is shown in the diagram. It is rotated 360° about the y-axis. Find the volume of the solid, as a multiple of π.

• \(y^{2}=x^{2}+9\Rightarrow x^{2}=y^{2}-9\)
• At \(x=0\), \(y=\sqrt{9}=3\). The region runs from \(y=3\) to \(y=5\).
• \(V=\pi\int_{3}^{5}\left(y^{2}-9\right)dy\)
• \(V=\pi\left[\frac{y^{3}}{3}-9y\right]_{3}^{5}=\frac{44\pi}{3}\)

A lamp shade is made by rotating a curve about the y-axis, with lengths in centimetres. The part of the curve \(y=\frac{x^{2}}{3}+1\) for \(0\le x\le 3\) is rotated 360° about the y-axis. Find the volume of the solid enclosed by the curve, the y-axis and the horizontal lines through the two ends of the curve.

• \(y=\frac{x^{2}}{3}+1\Rightarrow x^{2}=3\left(y-1\right)\)
• Limits: \(x=0\) gives \(y=1\), and \(x=3\) gives \(y=4\)
• \(V=\pi\int_{1}^{4}3\left(y-1\right)\,dy\)
• From the calculator: \(\pi\int_{1}^{4}3\left(y-1\right)\,dy = 42.4\)

A horn is made by rotating a curve about the y-axis, with lengths in decimetres. The part of the curve \(y=1+\ln x\) for \(1\le x\le 6\) is rotated 360° about the y-axis. Find the volume of the solid enclosed by the curve, the y-axis and the horizontal lines through the two ends of the curve.

• \(y=1+\ln x\Rightarrow \ln x=y-1\Rightarrow x=e^{y-1}\), so \(x^{2}=e^{2y-2}\)
• Limits: \(x=1\) gives \(y=1\), and \(x=6\) gives \(y=1+\ln 6\)
• \(V=\pi\int_{1}^{1+\ln 6}e^{2y-2}\,dy\)
• From the calculator: \(\pi\int_{1}^{1+\ln 6}e^{2y-2}\,dy = 55.0\)

A plant pot is made by rotating the curve \(y=12\ln(x+1)\), for \(0\le y\le 24\), 360° about the y-axis. Lengths are in cm.

1. (a) Find the capacity of the pot in litres.

   • \(y=12\ln(x+1)\Rightarrow x=e^{y/12}-1\)
   • \(V=\pi\int_0^{24}\left(e^{y/12}-1\right)^2dy\)
   • From the calculator: V = 604 cm³ = 0.604 litres

2. (b) The pot holds 400 cm³ of water. Find the depth of the water.

   • Solve \(\pi\int_0^{h}\left(e^{y/12}-1\right)^2dy=400\) for h
   • From the calculator: h = 22.1 cm
   • This is less than 24, so it fits

The diagram shows the region bounded by the curve \(x=6-(y-3)^{2}\) and the line \(x=2\). The region is rotated 360° about the y-axis. Find the volume of the solid, as a multiple of π.

• Curve meets line: \(6-(y-3)^{2}=2\Rightarrow (y-3)^{2}=4\), so \(y=1\) and \(y=5\)
• Outer radius \(x=6-(y-3)^{2}\) (the curve), inner radius \(x=2\) (the line)
• \(V=\pi\int_{1}^{5}\left[\left(6-(y-3)^{2}\right)^{2}-2^{2}\right]dy\)
• From the calculator: \(\pi\int_{1}^{5}\left[\left(6-(y-3)^{2}\right)^{2}-2^{2}\right]dy = 241\)
• Exact value: \(\frac{384\pi}{5}\)

A candle holder is made by rotating a curve about the y-axis. The part of the curve \(y=x^{2}+2\) for \(0\le x\le k\) is rotated 360° about the y-axis. The solid enclosed by the curve, the y-axis and the horizontal lines through the ends of the curve has volume 80 cm³. Find the value of k.

• \(x^{2}=y-2\). When \(x=0\), \(y=2\). When \(x=k\), \(y=k^{2}+2\).
• \(V=\pi\int_{2}^{k^{2}+2}(y-2)\,dy=\pi\left[\frac{(y-2)^{2}}{2}\right]_{2}^{k^{2}+2}=\frac{\pi k^{4}}{2}\)
• \(\frac{\pi k^{4}}{2}=80\Rightarrow k^{4}=50.9\)
• k = 2.67